Question

Y=x*arctanx
when does y'=0???

Answer 1

i believe the first step would be to differentiate =_= this sounds complicated...

Answer 2

is it y' =0 when x=0

Answer 3

\[\frac{d(x*\tan^{-1}x)}{dx} = \frac{x}{1 + x^2} + \tan^{-1}x\]

Answer 4

i wonder how we can solve for x
\[\large \frac{x}{1+ x^2} + \tan^{-1} x = 0\]

Answer 5

yup it is when x = 0 and no other solution

Answer 6

we all know that..but *how* do we prove it..

Answer 7

this is hard to prove i think..

Answer 8

take tan(inverse)x on other side

and draw graphs of both
arctan(x)
and -x/(1+x^2)
you will find their graphs cutting each other at x = 0 only

Answer 9

hmm i wonder if it's possible algebraically

Answer 10

i have to study the monotony ,asimptotes and then graph it-_- and i haven't found the roots yet :(

Answer 11

no not possible that way @igbasallote

Answer 12

yes @angela210793 u have to study that to solve these type of questions.

Answer 13

i know...but i can't find the roots :'(

Answer 14

so better study these topics first then find the roots

Answer 15

oh wait...(atctg)'=1/1+x^2 no?

Answer 16

Yes you are doing wrong earlier..

Answer 17

when y'=o?

Answer 18

\[x + \tan^{-1}x^2 + \tan^{-1}x = 0\]

\[x^2.\tan^{-1}x + x + \tan^{-1}x = 0\]

\[D = b^2 - 4ac\]

\[D = 1 - 4\tan^{-2}x\]

\[x = \frac{-1 \pm \sqrt{1-4\tan^{-2}x}}{2\tan^{-1}x}\]

Answer 19

Well, I am just trying..
Ha ha ha..

Answer 20

:O i dont understand anyway can u check if this is right...two different methods gave me two different answers :(